What You Should Already Know About Circuits
by Dr. Colton (Fall 2016)
We’re only going to spend a brief amount of time reviewing things that you should have learned in
Physics 220. Here’s a run-down of some of the things that you should already know. (Yes, you were
supposed to remember these things!) Review these items and if there are things you do not recall, review
the relevant sections of your old Phys 220 textbook. (Yes, you were supposed to keep your old textbook!)
1. Series and Parallel Resistors
The voltage (or more specifically, voltage difference of one side relative to the other) of a resistor is given
by Ohm’s law: ΔVR = IR.
A combination of 2 resistors can be labeled either as “series” or “parallel”, depending on how they are
connected.
SERIES
PARALLEL
Series resistors share the same current; parallel resistors share the same potential difference (V). The
equivalent resistance for two series and parallel resistors is given by these formulas:
SERIES:
Req = R1 + R2
PARALLEL:
(1)
or
Req = (1/R1 + 1/R2)-1
(2)
I will often use the symbol “//” to indicate a parallel resistance combination:
R1 // R2 = (1/R1 + 1/R2)-1
(definition of “//” symbol)
(3)
It can easily be proven that the series and parallel formulas extend as you would expect to any number of
resistors, such as Req = R1 + R2 + R3 + … for series resistors and Req = (1/R1 + 1/R2 + 1/R3 + …)-1 for
parallel resistors.
2. How to solve circuits with one battery and a mess of resistors
The series and parallel formulas are extremely powerful because networks of resistors can often be
decomposed into combinations of resistors that are in series and parallel with each other.
The way to tackle networks with many resistors is to slowly simplify the circuit by substituting in
equivalent resistors for combinations of series and/or parallel resistors. Then use repeated applications of
Ohm’s law to solve for the quantity you are looking for. For example, using “R12” to represent the
equivalent resistance of R1 and R2, etc., here’s how I would go about figuring out the equivalent resistance
of this network:
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Sample Problem 1: What’s the equivalent resistance of the 4 resistors?
R1
R3
R2
R4
R3
R12
R12 is given by parallel resistance
of R1 and R2, R12 = (1/R1 + 1/R2)-1
R4
R34 is given by parallel resistance
of R3 and R4, R34 = (1/R3 + 1/R4)-1
R12
R34
R1234 is given by series resistance of R12 and R34,
R1234 = (1/R1 + 1/R2)-1 + (1/R3 + 1/R4)-1
R1234
Shortcut notation:
R1234 = R1//R2 + R3//R4
Now suppose the four resistors were hooked up to a battery with voltage VB such that current flows
through the circuit from left to right. The total current flowing from the battery is Itot = VB/R1234. You
could then use that information to determine currents through other parts of the circuit.
Sample Problem 2: What is the current through R3?
One way to solve this problem would be to proceed as follows:
1. First, break down the whole circuit in terms of series and parallel equivalent resistances and solve
for the total current flowing from the battery (as already done).
2. Treating R1 and R2 as a single resistor through which Itot is flowing, Ohm’s Law says that the V
for R12 is equal to ItotR12. Thus the voltage just to the right of R1 and R2 (and just to the left of R3
and R4) must be: VB  ItotR12.
3. Because the right hand side of R3 is at 0 V, we can apply Ohm’s law to find the V across R3:
Δ
0
where Itot and R12 are as solved for above. Problem solved!
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To reiterate, the technique of using Ohm’s law for various parts of the circuit, together with the series and
parallel formulas, can be used to solve a number of circuit problems as long as you only have a bunch of
resistors together with a single battery.
3. Kirchhoff’s Laws
Kirchhoff’s two circuit laws are this: (KCL = Kirchhoff’s current law and KVL = Kirchhoff’s voltage
law)
KCL: ∑
KVL: ∑
0
Δ
(6)
0
(7)
Kirchhoff’s laws can frequently be used to solve circuit problems in situations that are too complicated
for series/parallel analysis such as done above.
Sample Problem 3: What is the current through the 500 Ω resistor?
100 
5V
200 
I1
I2
I3
500 
10 V
For problems like this, the typical method taught in physics books is to use Kirchhoff’s laws to obtain
simultaneous equations using the currents I1, I2, and I3 as unknowns. (Engineers learn a similar but
slightly quicker method called the “mesh current” method; we’ll discuss that in class.)
The “junction rule” and “loop rule” equations are:
I1 = I2 + I3
100 I1  500 I3 + 10 = 0
100 I1  5  200 I2 + 10 = 0
(junction equation)
(left hand loop)
(outer loop)
You could also do a loop rule equation for the right-hand loop, but it would be redundant with the left and
outer loop equations (it’s the outer loop equation minus the left hand loop equation, to be precise).
Once you have enough linearly independent simultaneous equations, solve them via your favorite method.
My own favorite method is to turn it into a matrix equation and use the matrix inverse to solve for the
“vector” of currents:
1
 1   I1   0 
 1


   
0
 500    I 2     10 
  100
  100  200
0   I 3    5 

matrix equivalent of the three equations
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1
1
1   0 
 I1   1
  
 

 500     10  multiply by the matrix inverse on both sides
0
 I 2     100
 I    100  200
0    5 
 3 
 I 1   0.026471... 
  

 I 2    0.011765... 
 I   0.014706... 
 3 

the answers
Here’s the code I’d use in Mathematica for that type of thing:
m = {{1, -1, -1}, {-100, 0, -500}, {-100, -200, 0}};
v = {0, -10, -5};
currents = Inverse[m].v // MatrixForm // N
If you have a very large number of simultaneous equations there are more efficient ways to solve them
than calculating the matrix inverse, but for anything we’d get in this class this method will work just fine.
4. Series and Parallel Capacitors
The voltage of a capacitor depends on its charge: ΔVC = Q/C.
When capacitors are connected together, they can be said to be in series or parallel just like resistors.
However, the formulas for the “equivalent capacitance” of the series/parallel combination are the reverse
of the resistor formulas.
SERIES:
PARALLEL:
or
Ceq = (1/C1 + 1/C2)-1
Ceq = C1 + C2
(4)
(5)
Circuits having just one battery and a network of capacitors can often be analyzed using only the series
and parallel formulas. This is very analogous to the resistor networks discussed above.
Due to the opposite nature of the parallel capacitors vs. parallel resistors formulas, I never use the symbol
“//” to indicate parallel capacitances. (I do, however, use it to indicate parallel impedances of capacitors,
which we’ll talk about in class.)
5. Time dependence of simple RC circuits
When a capacitor is discharged through a resistor, the charge on the capacitor (and hence the voltage of
the capacitor) decays exponentially.
/
Δ
Δ
/
discharging capacitor
(8)
discharging capacitor
(9)
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where Q0 and ΔV0 (= Q0/C) are the capacitor’s initial charge and voltage respectively, and the “time
constant” τ is given by:
(10)
After a time equal to τ, the charge on the capacitor is equal to 36.8% ( e-1) of its initial value.
When an initially uncharged capacitor is connected to a battery, similar equations hold:
/
1
Δ
Δ
1
/
charging capacitor
(11)
charging capacitor
(12)
where Qmax and ΔVmax (= Qmax/C) represent the final charge and voltage of the capacitor, respectively, and
τ = RC is the same time constant as for discharging capacitors. After a time equal to τ, the charge on the
capacitor is equal to 63.2% ( 1 – e-1) of its maximum value. The reason for this nonlinear charging is that
the more “full” the capacitor is, the harder it gets to add more charge.
Proof of the time dependent RC equations. Eqs. (8) and (11) can be proved via KVL and the “guess and
check” method of solving differential equations, and Eqs. (9) and (12) follow by the voltage of a capacitor
ΔVC = Q/C. For example, for a charging series RC circuit, KVL yields:
0
which by using I = dQ/dt can be turned into this:
0
If we guess
1
/
, we find
/
/
1
1
0
/
0
and
. Moreover our guess satisfies the initial condition
That equation is true if
of
0
0. Therefore by the uniqueness theorem of first order differential equations, Eq. (11) is the
solution for this situation. (Eq. 8 is proved similarly for a discharging capacitor.)

6. Inductors
The voltage of an inductor depends on the rate of change of its current: ΔVL = –L dI/dt. The negative sign
follows from Lenz’s law. Inductors can be thought of as current-storing devices, similar to how capacitors
are charge-storing devices.
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Inductors use the same series and parallel formulas as resistors, with the caveat that if two inductors are
placed too near each other, their mutual inductance must also be taken into account.
7. Time dependence of simple RL circuits
When the current stored in an inductor is discharged through a resistor, the current and the voltage of the
inductor both decay exponentially.
/
Δ
/
Δ
discharging inductor
(13)
discharging inductor
(14)
where I0 and ΔV0 (= I0L/τ) are the inductor’s initial current and voltage respectively, and the “time
constant” τ is given by:
/ (15)
After a time equal to τ, the current through the inductor is equal to 36.8% ( e-1) of its initial value.
When an inductor is connected to a battery with no initial current, similar equations hold:
/
1
/
Δ
Δ
charging inductor
(16)
charging inductor
(17)
where Imax and ΔVmax (=ImaxL/τ) represent the final, maximum current and the maximum voltage
magnitude (which incidentally occurs at t = 0) of the inductor, respectively. The time constant τ = L/R is
the same as for discharging inductors. After a time equal to , the current through the inductor is equal to
63.2% ( 1 – e-1) of its maximum value. The reason for this nonlinear charging is that the more “full” the
inductor is in terms of allowable current, the harder it gets to add more current.
Proof of the time dependent RL equations. Eqs. (13) and (16) can be proved via KVL and the “guess and
check” method of solving differential equations, and Eqs. (14) and (17) follow by the voltage of an
inductor ΔVL = –LdI/dt. For example, for a charging series RL circuit, KVL yields:
0
If we guess
1
/
, we find
/
1
/
0
0
/ and
/ . Moreover our guess satisfies the initial
That equation is true if
condition of
0
0. Therefore by the uniqueness theorem of first order differential equations,
Eq. (16) is the solution for this situation. (Eq. 13 is proved similarly for a discharging inductor.)
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